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Say that $|\sqrt{x}-1| < \epsilon$. I am having a problem with handling this inequality. I want to exclude x.

I.

$|\sqrt{x}-1| < \epsilon$

$|\sqrt{x}| - |1| \leq |\sqrt{x}-1| < \epsilon$

$|\sqrt{x}| - |1| < \epsilon$

$|\sqrt{x}| < \epsilon + |1|$

$|\sqrt{x}| < \epsilon + 1$

$-(\epsilon + 1)< \sqrt{x} < \epsilon + 1$

$(-(\epsilon + 1))^2< x < (\epsilon + 1)^2$

If however I try this:

II.

$|\sqrt{x}-1| < \epsilon$

$-\epsilon < \sqrt{x}-1 < \epsilon$

$1 - \epsilon < \sqrt{x} < 1+ \epsilon$

$(1-\epsilon)^2 < x < (1+\epsilon)^2$

What am I doing wrong in I. or II.?

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In $I$, the first step is incorrect : $$ |\sqrt{x} - 1| <\epsilon $$ is not equivalent to $$ |\sqrt{x}| - 1 < \epsilon $$ $II$ is correct though.

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